movement of the solid, EN comes into coincidence with FO, HQ with GP, and AL with BM. Hence, the solids A' and B' may be made to coincide throughout and are therefore equal. Now, from these equal solids A' and B' take away the common part BGFD-LQNJ and there remains AF equivalent to LO. Q. E. D.. Since the above demonstration was made wholly independent of the form of the base or right section involved, it applies as well to any other prism as to the parallelopiped. Hence, 77. COROLLARY 1. An oblique prism is equivalent to that right prism whose base and altitude are equal respectively to a right section and a lateral edge of the oblique prism. 78. COROLLARY 2. If a cuboid and a parallelopiped have equivalent bases and equal altitudes, they are equal to each other in volume. Produce the edges LQ, JN, MP, and KO; lay off OS equal to KO; and pass through O and S planes perpendicular to OS. Then, VR thus formed is a cuboid; and, since NV is a right section of LO, and OS equals KO, LO is equivalent to VR (76). But, AF is equivalent to LO, as we have shown (76); hence, VR is equivalent to AF. 79. COROLLARY 3. The volume of any parallelopiped is equal to the product of its base and altitude. Regard CF, JO, and NS, as the bases of the parallelopipeds to which they belong. Represent the common altitude of the three solids by a .. (25); then will the volume of AF be equal to (a) (CF). For CF, JO, and NS are equivalent (363); then, multiplying each of the above by a, we obtain (a) (CF), (a) (JO), and (a) (NS) equivalent to one another. But (a) (NS) equals the volume of VR (75); and the volume of VR equals the volume of AF (78). Hence, the volume of AF equals (a) (CF). Q. E. D. Note. We shall hereafter use the symbol to represent equivalence. PROPOSITION XVI. 80. THEOREM. A plane passed through the diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms. D' B' H F D B Let the plane AC' pass through the diagonally opposite edges AA' and CC' of the parallelopiped BD'; then will the triangular prisms ABC-B' and ADC-D', thus formed, be equivalent. Any right section EFGH is a parallelogram (66, b), and EG, drawn from vertex E to vertex G, is a diagonal. Hence the triangle EFG equals the triangle EHG. (359.) The prisms ABC-B′ and ADC-D' are equivalent, respectively, to those right prisms whose bases are the right sections EFG and EHG, and whose common altitude is the edge CC' (77). But EFG and EHG are equal; therefore the right prisms of which they are bases, are equal in vol ume (65). Hence the prism ABC-B' is equivalent to the prism ADC-D'. Q. E. D. 81. SCHOLIUM. ABC-B'and ADC-D' are mutually equal in all their parts; however, not being superposable, they are not equal solids but symmetrical (41 and 42). 82. COROLLARY. The volume of any triangular prism is equal to the product of its base and altitude. Let a represent the altitude of ABC-B'; then will the volume of ABC-B' be equal to (ABC)(a). (1). For, vol. BD' = (2)(ABC-B'). . . (80), (2). And, area DB = (2)(ABC)... (359), (3). But, vol. BD' Now substitute in equation (3), the values of vol. BD' and area DB from equations (1) and (2) respectively, and we find, (2) (ABC-B') = (2) (ABC)(a). Hence, ABC-B' = (ABC) (a). Q. E. D. DEFINITIONS. 83. A cylindrical surface is a curved surface generated by a line moving parallel to a given fixed line and continually touching a guiding curve, the curve and the fixed line not lying in the same plane. The generating line in any position is termed an element of the surface. N F B M D E CD is a moving line always parallel to NM a given fixed line, and continually touching DEA, a guiding curve. The guiding curve and the given fixed line lie in different planes. CD, FE, and BA are elements of the cylindrical surface DB. 84. A cylinder is a solid bounded by a cylindrical surface and two parallel planes called bases. The bases cut all the elements of the cylindrical surface. If the bases are circles as in the figure, the cylinder is said to be circular; and if the generating line is perpendicular to the planes of the circles, the cylinder is termed a right circular cylinder. The axis of the cylinder is the line joining the centres of the bases. оо 85. A right circular cylinder is called a cylinder of revolution; for if a rectangle be revolved about one of its sides as an axis, it will generate a right circular cylinder. 86. If two similar rectangles be revolved about homologous sides as axes, two similar cylinders of revolution will be generated. 87. A prism is inscribed in a cylinder when the bases of the prism are inscribed in the bases of the cylinder. The B lateral edges of the prism then become elements of the surface of the cylinder. FC is a prism inscribed in the cylinder FC. 88. A cylinder may be regarded as the limiting case of an inscribed prism, when the number of lateral faces of the prism is increased indefinitely. Thus if an inscribed prism of four lateral faces be changed to an inscribed prism of eight lateral faces, and then the prism of eight faces be again changed to one of sixteen faces, etc., the prism will approach the cylinder as a limit. By indefinitely increasing the number of faces of the inscribed prism, it may ultimately be brought into coin cidence with the cylinder, in altitude, in surface, and in volume; that is, in its entirety. 89. For definitions of altitude, right section, etc., of cylinders, see corresponding definitions under prisms. PROPOSITION XVII. 90. THEOREM. The volume of any prism is equal to the product of its base and altitude. Let the base, altitude, and volume of any prism DG, be represented by B, a, and V, respectively; then will V = (B) (a). Through the edge AB and the vertices E, K, etc., pass planes AF, AL, etc., dividing the prism DG into triangular prisms. All the prisms thus constructed have the common altitude a. From (82) we may write the following equations: vol. (DFB-C) = (DFB) (a). vol. (FBL-A = (FBL) (a). |